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Dima020 [189]
1 year ago
8

how many moles of sodium hydroxide would have to be added to 250 ml of a 0.403 m acetic acid solution, in order to prepare a buf

fer with a ph of 4.750?
Chemistry
1 answer:
jonny [76]1 year ago
6 0

The volume of the buffer solution having a ph value is calculated by henderson's hasselbalch equation.

Buffer solution is water based solution which consists of a mixture containing a weak acid and a conjugate base of the weak acid. or a weak base and conjugate acid of a weak base.it is a mixture of weak acid and a base. The pH of the buffer solution is determined by the expression of the henderson hasselbalch equation.

              pH=pKa + log [salt]/[acid]

Where, pKa =dissociation constant , A- = concentration of the conjugate base, [HA]= concentration of the acid. Here, a buffer solution contains 0.403m acetic acid  and 250 ml is added  in order to prepare a buffer with a ph of 4.750. Putting all the values in the henderson hasselbalch equation we find the pH of the buffer solution.

To learn more about hendersons hasselbalch equation please visit:

brainly.com/question/13423434

#SPJ4

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Identify limiting and excess reagents when 25g of nitrogen reacts with 25g of hydrogen. How many grams of ammonia gas are formed
Makovka662 [10]

Answer:

Nitrogen is limiting reactant while hydrogen is in excess.

Explanation:

Given data:

Mass of N₂ = 25 g

Mass of H₂ = 25 g

Mass of ammonia formed = ?

Solution:

Chemical equation:

N₂ + 3H₂    →     2NH₃

Number of moles of Nitrogen:

Number of moles = mass/ molar mass

Number of moles = 25 g/ 28 g/mol

Number of moles = 0.89 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 25 g/ 2 g/mol

Number of moles = 12.5 mol        

Now we will compare the moles of both reactant with ammonia.

                   H₂            ;             NH₃

                    3             :              2

                    12.5        :            2/3×12.5 = 8.3

                 

                   N₂            ;             NH₃

                    1              :              2

                    0.89        :            2×0.89 = 1.78

The number of moles of ammonia produced by nitrogen are less thus nitrogen is limiting reactant while hydrogen is in excess.

7 0
4 years ago
The value of ΔG° at 221.0°C for the formation of phosphorous trichloride from its constituent elements, P2(g) + 3Cl2(g) → 2PCl3(
densk [106]

Answer:

ΔG    = - 590.20 kJ/mol

Explanation:

The formula for calculating Gibb's Free Energy  can be written as:

ΔG    =    ΔH - TΔS

Given That:

ΔH = -720.5 kJ/mol

T = 221.0°C = (221.0 + 273.15) = 494.15 K

ΔS° = -263.7 J/K

So; ΔS° = -0.2637 kJ/K if being converted from joule to Kilo-joule

Since we are all set, let replace our given data in the above equation:

ΔG    =  (-720.5 kJ/mol) - (494.15 K) ( - 0.2637 kJ/K)

ΔG    =  (-720.5 kJ/mol) - (- 130.30755)

ΔG    =  -720.5 kJ/mol + 130.30755

ΔG    =  -590.192645 kJ/mol

ΔG    = - 590.20 kJ/mol

Thus, The value of ΔG° at 221.0°C for the formation of phosphorous trichloride from its constituent elements, P2(g) + 3Cl2(g) → 2PCl3(g) is <u>-590.20</u> kJ/mol.

3 0
3 years ago
The metric system is only used in France<br><br> True<br> or<br> False
Klio2033 [76]
False. Most countries (not USA) use the metric system. England being one of them
4 0
3 years ago
. What is the main gas found in the air we breathe?
laiz [17]

Answer:

Nitrogen

Explanation:

4 0
4 years ago
1. explain how you would make 100 mL of a 2.0% (w/v) solution of sodium chloride?
sleet_krkn [62]

1.

Percent (mass/volume) of the solution = (mass of the solute / volume of the solution) x 100

Given, percent of NaCl (m/v) solution = 2.0 %

Volume of solution = 100 ml

Plugging in the numbers in the formula we get,

2.0 = (mass of NaCl/ 100) x 100

Mass of NaCl = (2.0 x 100)/100 = 2 g

So we would dissolve 2g NaCl(s) in 100 ml water to form a 2.0% NaCl solution.

2.

Percent (mass/volume) of the solution = (mass of the solute / volume of the solution) x 100

Given, percent of MgSO₄ (m/v) solution = 1.6%

Volume of solution = 250 ml

Plugging in the numbers in the formula we get,

1.6 = (mass of MgSO₄/ 250) x 100

Mass of MgSO₄ = (1.6 x 250)/100 = 4 g

3.

Percent (mass/volume) of the solution = (mass of the solute / volume of the solution) x 100

Given, mass of CuSO₄ = 2.7 g

Volume of solution = 75 ml

Percent (mass/volume) of the solution = (2.7 g/ 75 ml) x 100 = 3.6%

7 0
3 years ago
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